Chapter 4 Transients - Department of Electrical and ... Notes/Chapter 04.pdf · ELECTRICAL...
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ELECTRICAL ENGINEERING Principles and Applications SE OND EDITION Chapter 4 Transients Chapter 4 Transients 1. Solve first-order RC or RL circuits. 2. Understand the concepts of transient response and steady-state response.
Transcript of Chapter 4 Transients - Department of Electrical and ... Notes/Chapter 04.pdf · ELECTRICAL...
ELECTRICAL
ENGINEERINGPrinciples and
Applications
SE OND EDITION
Chapter 4Transients
Chapter 4Transients
1. Solve first-order RC or RL circuits.
2. Understand the concepts of transient response and steady-state response.
ELECTRICAL
ENGINEERINGPrinciples and
Applications
SE OND EDITION
Chapter 4Transients
3. Relate the transient response of first-order circuits to the time constant.
4. Solve RLC circuits in dc steady-state conditions.
5. Solve second-order circuits.
6. Relate the step response of a second-order systemto its natural frequency and damping ratio.
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
Transients
The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integrodifferential equations.
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
Discharge of a Capacitance through a Resistance
( ) ( ) 0=+R
tvdt
tdvC CC
( ) ( ) 0=+ tvdt
tdvRC CC
( ) stC Ketv =
0=+ stst KeRCKse
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
RCs 1−=
( ) RCtC Ketv −= ( ) RCt
iC eVtv −=
( ) iC Vv =+0
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
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Chapter 4Transients
The time interval τ = RC iscalled the time constant of
the circuit.
( ) τtssC eVVtv −−=
ELECTRICAL
ENGINEERINGPrinciples and
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Chapter 4Transients
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
DC STEADY STATE
The steps in determining the forced response for RLC circuits with dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
RL CIRCUITS
The steps involved in solving simple circuits containing dc sources, resistances, and one energy-storage element (inductance or capacitance) are:
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
1. Apply Kirchhoff’s current and voltage laws to write the circuit equation.
2. If the equation contains integrals, differentiate each term in the equation to produce a pure differential equation.
3. Assume a solution of the form K1 + K2est.
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
4. Substitute the solution into the differential equation to determine the values of K1 and s . (Alternatively, we can determine K1 by solving the circuit in steady state as discussed in Section 4.2.)
5. Use the initial conditions to determine the value of K2.
6. Write the final solution.
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
RL Transient Analysis
( ) LtReKti −+= 22
Time constant is
RL=τ
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
Applications
SE OND EDITION
Chapter 4Transients
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
RC AND RL CIRCUITS WITH GENERAL SOURCES
The general solution consists of two parts.
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Chapter 4Transients
The particular solution (also called the forced response) is any expression that satisfies the equation.
In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution.
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Chapter 4Transients
The homogeneous equation is obtained by setting the forcing function to zero.
The complementary solution (also called the natural response) is obtained by solving the homogeneous equation.
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Chapter 4Transients
Step-by-Step Solution
Circuits containing a resistance, a source,and an inductance (or a capacitance)
1. Write the circuit equation and reduce it to a first-order differential equation.
ELECTRICAL
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Chapter 4Transients
2. Find a particular solution. The details of this step depend on the form of the forcing function. We illustrate several types of forcing functions in examples, exercises, and problems.
3. Obtain the complete solution by adding the particular solution to the complementarysolution given by Equation 4.44, which contains the arbitrary constant K.
4. Use initial conditions to find the value of K.
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
ENGINEERINGPrinciples and
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SE OND EDITION
Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
SECOND-ORDER CIRCUITS
( ) ( ) ( ) ( ) ( )tvvdttiC
tRidt
tdiL sC
t
=+++ ∫ 01
0
LR
2=α
LC1
0 =ω
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SE OND EDITION
Chapter 4Transients
( ) ( ) ( ) ( )tftidt
tdidt
tid =++ 202
2
2 ωα
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Chapter 4Transients
0ωαζ =
20
21 ωαα −+−=s
20
22 ωαα −−−=s
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Chapter 4Transients
1. Overdamped case (ζ > 1). If ζ > 1 (or equivalently, if α > ω0), the roots of the characteristic equation are real and distinct. Then the complementary solution is
( ) tstsc eKeKtx 21
21 +=
In this case, we say that the circuit is overdamped.
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Chapter 4Transients
2. Critically damped case (ζ = 1). If ζ = 1 (or equivalently, if α = ω0 ), the roots are real and equal. Then the complementary solution is
( ) tstsc teKeKtx 11
21 +=
In this case, we say that the circuit is critically damped.
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Chapter 4Transients
3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently, if α < ω0), the roots are complex. (By the term complex, we mean that the roots involve the square root of –1.) In other words, the roots are of the form
nn jsjs ωαωα −−=+−= 21 and
in which j is the square root of -1 and the natural frequency is given by
220 αωω −=n
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Chapter 4Transients
In electrical engineering, we use j rather than i to stand for square root of -1, because we use i for current.
For complex roots, the complementary solution is of the form
( ) ( ) ( )teKteKtx nt
nt
c ωω αα sincos 21−− +=
In this case, we say that the circuit is underdamped.
ELECTRICAL
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Chapter 4Transients
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Chapter 4Transients
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Chapter 4Transients
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Chapter 4Transients
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Chapter 4Transients
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Chapter 4Transients
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Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
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Chapter 4Transients
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SE OND EDITION
Chapter 4Transients
ELECTRICAL
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SE OND EDITION
Chapter 4Transients
